Answer :
To determine the length of a pendulum required for it to have a period of 1 second per cycle on the moon, where the acceleration due to gravity is \( g_{\text{moon}} = 1.6 \, \text{N/kg} \), we will use the formula that relates the period of a pendulum to its length and the acceleration due to gravity.
The formula for the period \( T \) of a simple pendulum is given by:
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
Where:
- \( T \) is the period of the pendulum (in seconds).
- \( L \) is the length of the pendulum (in meters).
- \( g \) is the acceleration due to gravity (in N/kg).
We are given:
- The desired period \( T = 1 \) second.
- The acceleration due to gravity on the moon \( g = 1.6 \, \text{N/kg} \).
We need to calculate the length \( L \) for the pendulum to have this period. First, we solve for \( L \):
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
We rearrange the formula to solve for \( L \):
[tex]\[ \sqrt{\frac{L}{g}} = \frac{T}{2\pi} \][/tex]
Square both sides to eliminate the square root:
[tex]\[ \frac{L}{g} = \left(\frac{T}{2\pi}\right)^2 \][/tex]
Multiply both sides by \( g \):
[tex]\[ L = g \left(\frac{T}{2\pi}\right)^2 \][/tex]
Substitute the given values \( T = 1 \, \text{s} \) and \( g = 1.6 \, \text{N/kg} \):
[tex]\[ L = 1.6 \left(\frac{1}{2\pi}\right)^2 \][/tex]
This simplifies to:
[tex]\[ L = 1.6 \left(\frac{1}{6.2832}\right)^2 \][/tex]
Calculating the fraction inside the parentheses:
[tex]\[ L = 1.6 \left(\frac{1}{6.2832}\right)^2 \approx 1.6 \times 0.02533 \approx 0.040528 \][/tex]
Thus, the length \( L \) of the pendulum must be approximately \( 0.040528 \) meters.
Comparing this result with the given choices:
- \( 3.2 \, \text{m} \)
- \( 7.9 \, \text{m} \)
- \( 0.25 \, \text{m} \)
- \( 0.041 \, \text{m} \)
The closest choice to our calculated length of \( 0.040528 \, \text{m} \) is:
[tex]\[ 0.041 \, \text{m} \][/tex]
Therefore, the correct length of the pendulum for it to have a period of 1 second per cycle on the moon is [tex]\( 0.041 \, \text{m} \)[/tex].
The formula for the period \( T \) of a simple pendulum is given by:
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
Where:
- \( T \) is the period of the pendulum (in seconds).
- \( L \) is the length of the pendulum (in meters).
- \( g \) is the acceleration due to gravity (in N/kg).
We are given:
- The desired period \( T = 1 \) second.
- The acceleration due to gravity on the moon \( g = 1.6 \, \text{N/kg} \).
We need to calculate the length \( L \) for the pendulum to have this period. First, we solve for \( L \):
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
We rearrange the formula to solve for \( L \):
[tex]\[ \sqrt{\frac{L}{g}} = \frac{T}{2\pi} \][/tex]
Square both sides to eliminate the square root:
[tex]\[ \frac{L}{g} = \left(\frac{T}{2\pi}\right)^2 \][/tex]
Multiply both sides by \( g \):
[tex]\[ L = g \left(\frac{T}{2\pi}\right)^2 \][/tex]
Substitute the given values \( T = 1 \, \text{s} \) and \( g = 1.6 \, \text{N/kg} \):
[tex]\[ L = 1.6 \left(\frac{1}{2\pi}\right)^2 \][/tex]
This simplifies to:
[tex]\[ L = 1.6 \left(\frac{1}{6.2832}\right)^2 \][/tex]
Calculating the fraction inside the parentheses:
[tex]\[ L = 1.6 \left(\frac{1}{6.2832}\right)^2 \approx 1.6 \times 0.02533 \approx 0.040528 \][/tex]
Thus, the length \( L \) of the pendulum must be approximately \( 0.040528 \) meters.
Comparing this result with the given choices:
- \( 3.2 \, \text{m} \)
- \( 7.9 \, \text{m} \)
- \( 0.25 \, \text{m} \)
- \( 0.041 \, \text{m} \)
The closest choice to our calculated length of \( 0.040528 \, \text{m} \) is:
[tex]\[ 0.041 \, \text{m} \][/tex]
Therefore, the correct length of the pendulum for it to have a period of 1 second per cycle on the moon is [tex]\( 0.041 \, \text{m} \)[/tex].
